Let A = \begin{bmatrix} a & 1 \\ -1 & b \end{bmatrix} where a and b are real number. If A^2 is a null matrix then the product ab equals-
Step-by-Step Solution
Key Concept: A^2 = 0 implies the characteristic equation of A is \lambda^2 = 0. By Cayley-Hamilton theorem, A^2 - tr(A)A + det(A)I = 0. Since A^2 = 0, we must have tr(A) = 0 and det(A) = 0.
Given A = \begin{bmatrix} a & 1 \\ -1 & b \end{bmatrix}. A^2 = \begin{bmatrix} a & 1 \\ -1 & b \end{bmatrix} \begin{bmatrix} a & 1 \\ -1 & b \end{bmatrix} = \begin{bmatrix} a^2-1 & a+b \\ -a-b & b^2-1 \end{bmatrix}. Since A^2 is a null matrix, a^2-1=0, b^2-1=0, and a+b=0. From a+b=0, b=-a. Substituting into a^2-1=0 gives a^2=1, so a=1 or a=-1. If a=1, b=-1, then ab = -1. If a=-1, b=1, then ab = -1.
Correct Answer: 3